The final temperature.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

a)

Expert Solution

Answer to Problem 215RP

The final temperature is 315.3 K.

Explanation of Solution

Write the expression for the energy balance equation for the closed system.

ΔEin−ΔEout=ΔEsystem (I)

Here, energy change in to the system is ΔEin, energy change exit from the system is ΔEout and change in energy of the system is ΔEsystem.

Write the expression for the initial pressure in the cylinder using ideal gas law.

P1=m1RT1ν1 (II)

Here, initial pressure is P1, initial mass is m1, gas constant is R, initial temperature is T1 and initial volume is ν1.

Write the expression for the mass of the air at final stage using ideal gas law.

m2=P2ν2RT2 (III)

Here, final mass is m2, final pressure is P2, final specific volume is ν2, gas constant is R, and final temperature is T2.

Write the expression for the mass entering the cylinder (mi).

mi=m2−m1 (IV)

Conclusion:

Substitute mihi for ΔEin, Wb,out for ΔEout and m2u2−m1u1 for ΔEsystem in Equation (I).

mihi−Wb,out=m2u2−m1u1micpTi−Wb,out=m2cvT2−m1cvT1 (V)

Here, work done during the process is Wb,out.mass flow is m, specific heat at constant pressure is cp , specific heat at constant volume is cv, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Substitute 1.3 kg for m1, 0.287 kJ/kg⋅K for R, 30°C for T1, and 0.40 m3 for ν1 in Equation (II).

P1=(1.3 kg)(0.287 kJ/kg⋅K)30°C0.40 m3=(1.3 kg)(0.287 kJ/kg⋅K)(30+273) K0.40 m3=282.6 kPa

When system undergoes constant pressure process, P2=P1

Substitute 282.6 kPa for P2, 1.5×0.40 m3 for ν2, 0.287 kJ/kg⋅K for R in Equation (III).

m2=(282.6 kPa)(1.5×0.40 m3)(0.287 kJ/kg⋅K)T2=590.8T2 (VI)

Substitute 590.8T2 for m2 and 1.3 kg for m1 in Equation (IV).

mi=590.8T2−1.3 kg=590.8T2−1.3 (VII)

Substitute 590.8T2−1.3 for mi, 1.005 kJ/kg⋅K for cp, 70°C for Ti, 56.52 kJ for Wb,out, 590.8T2 for m2, 0.718 kJ/kg⋅K for cv, 1.3 kg for m1 and 30°C for T1 in Equation (V).

{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70°C)−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)(30°C)}{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70+273)K−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)×(30+273)K}

(590.8T2−1.3)344.715−56.52=424.19−282.82203657.6T2−504.65=141.3742T2=203657.6646.02 =315.3 K

Thus, the final temperature is 315.3 K.

b)

To determine

The amount of mass entered the cylinder.

b)

Expert Solution

Answer to Problem 215RP

The amount of mass entered the cylinder is 0.5742 kg.

Explanation of Solution

Substitute 315.3 K for T2 in Equation (VII).

m2=590.8315.3 K=1.874 kg

Substitute 1.874 kg for m2 and 1.3 kg for m1 in Equation (V).

mi=1.874 kg−1.3 kg= 0.5742 kg

Thus, the amount of mass entered the cylinder is 0.5742 kg.

c)

To determine

The work done during the process.

c)

Expert Solution

Answer to Problem 215RP

The work done during the process is 56.52 kJ.

Explanation of Solution

Write the expression for the work done during the process (Wb,out).

Wb,out=P1(ν2−ν1) (VIII)

Here, initial volume is ν1 , final volume is ν2 and initial pressure is (P1).

Conclusion:

Substitute 282.6 kPa for P1, (150 %×0.40 m3) for ν2, and 0.40 m3 for ν1 in Equation (VIII).

Wb,out=282.6 kPa(1.5×0.40 m3−0.40 m3)=56.52 kJ

Thus, the work done during the process is 56.52 kJ.

d)

To determine

The entropy generated for the process.

d)

Expert Solution

Answer to Problem 215RP

The entropy generated for the process is 0.09714 kJ/K.

Explanation of Solution

Write the expression for the entropy generated during the process.

Sgen=m2s2−m1s1−misi=m2s2−m1s1−(m2−m1)si=m2(s2−si)−m1(s1−si)=m2(cplnT2Ti−RlnP2Pi)−m1(cplnT1Ti−RlnP1Pi) (IX)

Here, entropy generation is Sgen, initial entropy is s1,entropy entering the cylinder is si , final entropy is s2 , pressure entering the cylinder is Pi , and final pressure is P2 , temperature entering the cylinder is Ti , final temperature is T2 , gas constant is R , specific heat at constant pressure is cp, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

Conclusion:

Substitute 1.874 kg for m2, 1.005 kJ/kg⋅K for cp, 315.3 K for T2, 70 C for Ti, 0.287 kJ/kg⋅K for R, 282.6 kPa for P2, 500 kPa for Pi, 1.3 kg for m1, 30 C for T1, and 282.6 kPa for P1 in Equation (IX).

Sgen={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln30°C70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln(30+273) K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}=0.09714 kJ/K

Thus, the entropy generated for the process is 0.09714 kJ/K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Q2/ A mass of 2.4 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now compressed to a final pressure of 600 kPa. During the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during this process. Answer: 272 Kj

Refrigerant-134a enters the compressor of a refrigeration system as saturated vapor at 0.14 MPa and leaves as superheated vapor at 0.8 MPa and 60°C at a rate of 0.06 kg/s. Determine the rates of energy transfers by mass into and out of the compressor. Assume the kinetic and potential energies to be negligible. The enthalpies of refrigerant-134a at the inlet and the exit are 239.19 kJ/kg and 296.82 kJ/kg, respectively. The rate of energy transfer by mass into the compressor is The rate of energy transfer by mass out of the compressor is kW. kW.

A piston-cylinder device with a set of stops initially contains 0.5 kg of steam at 1.2 MPa and 500°C. The location of the stops corresponds to 30 percent of the initial volume. Now the steam is cooled. Determine the compression work if the final state is;(a) 1.2 MPa and 200°C.(b) 500 kPa(c) Also determine the temperature at the final state in part (b).

Answer 2

Question

Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

a)

Expert Solution

Answer to Problem 215RP

The final temperature is 315.3 K.

Explanation of Solution

Write the expression for the energy balance equation for the closed system.

ΔEin−ΔEout=ΔEsystem (I)

Here, energy change in to the system is ΔEin, energy change exit from the system is ΔEout and change in energy of the system is ΔEsystem.

Write the expression for the initial pressure in the cylinder using ideal gas law.

P1=m1RT1ν1 (II)

Here, initial pressure is P1, initial mass is m1, gas constant is R, initial temperature is T1 and initial volume is ν1.

Write the expression for the mass of the air at final stage using ideal gas law.

m2=P2ν2RT2 (III)

Here, final mass is m2, final pressure is P2, final specific volume is ν2, gas constant is R, and final temperature is T2.

Write the expression for the mass entering the cylinder (mi).

mi=m2−m1 (IV)

Conclusion:

Substitute mihi for ΔEin, Wb,out for ΔEout and m2u2−m1u1 for ΔEsystem in Equation (I).

mihi−Wb,out=m2u2−m1u1micpTi−Wb,out=m2cvT2−m1cvT1 (V)

Here, work done during the process is Wb,out.mass flow is m, specific heat at constant pressure is cp , specific heat at constant volume is cv, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Substitute 1.3 kg for m1, 0.287 kJ/kg⋅K for R, 30°C for T1, and 0.40 m3 for ν1 in Equation (II).

P1=(1.3 kg)(0.287 kJ/kg⋅K)30°C0.40 m3=(1.3 kg)(0.287 kJ/kg⋅K)(30+273) K0.40 m3=282.6 kPa

When system undergoes constant pressure process, P2=P1

Substitute 282.6 kPa for P2, 1.5×0.40 m3 for ν2, 0.287 kJ/kg⋅K for R in Equation (III).

m2=(282.6 kPa)(1.5×0.40 m3)(0.287 kJ/kg⋅K)T2=590.8T2 (VI)

Substitute 590.8T2 for m2 and 1.3 kg for m1 in Equation (IV).

mi=590.8T2−1.3 kg=590.8T2−1.3 (VII)

Substitute 590.8T2−1.3 for mi, 1.005 kJ/kg⋅K for cp, 70°C for Ti, 56.52 kJ for Wb,out, 590.8T2 for m2, 0.718 kJ/kg⋅K for cv, 1.3 kg for m1 and 30°C for T1 in Equation (V).

{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70°C)−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)(30°C)}{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70+273)K−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)×(30+273)K}

(590.8T2−1.3)344.715−56.52=424.19−282.82203657.6T2−504.65=141.3742T2=203657.6646.02 =315.3 K

Thus, the final temperature is 315.3 K.

b)

To determine

The amount of mass entered the cylinder.

b)

Expert Solution

Answer to Problem 215RP

The amount of mass entered the cylinder is 0.5742 kg.

Explanation of Solution

Substitute 315.3 K for T2 in Equation (VII).

m2=590.8315.3 K=1.874 kg

Substitute 1.874 kg for m2 and 1.3 kg for m1 in Equation (V).

mi=1.874 kg−1.3 kg= 0.5742 kg

Thus, the amount of mass entered the cylinder is 0.5742 kg.

c)

To determine

The work done during the process.

c)

Expert Solution

Answer to Problem 215RP

The work done during the process is 56.52 kJ.

Explanation of Solution

Write the expression for the work done during the process (Wb,out).

Wb,out=P1(ν2−ν1) (VIII)

Here, initial volume is ν1 , final volume is ν2 and initial pressure is (P1).

Conclusion:

Substitute 282.6 kPa for P1, (150 %×0.40 m3) for ν2, and 0.40 m3 for ν1 in Equation (VIII).

Wb,out=282.6 kPa(1.5×0.40 m3−0.40 m3)=56.52 kJ

Thus, the work done during the process is 56.52 kJ.

d)

To determine

The entropy generated for the process.

d)

Expert Solution

Answer to Problem 215RP

The entropy generated for the process is 0.09714 kJ/K.

Explanation of Solution

Write the expression for the entropy generated during the process.

Sgen=m2s2−m1s1−misi=m2s2−m1s1−(m2−m1)si=m2(s2−si)−m1(s1−si)=m2(cplnT2Ti−RlnP2Pi)−m1(cplnT1Ti−RlnP1Pi) (IX)

Here, entropy generation is Sgen, initial entropy is s1,entropy entering the cylinder is si , final entropy is s2 , pressure entering the cylinder is Pi , and final pressure is P2 , temperature entering the cylinder is Ti , final temperature is T2 , gas constant is R , specific heat at constant pressure is cp, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

Conclusion:

Substitute 1.874 kg for m2, 1.005 kJ/kg⋅K for cp, 315.3 K for T2, 70 C for Ti, 0.287 kJ/kg⋅K for R, 282.6 kPa for P2, 500 kPa for Pi, 1.3 kg for m1, 30 C for T1, and 282.6 kPa for P1 in Equation (IX).

Sgen={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln30°C70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln(30+273) K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}=0.09714 kJ/K

Thus, the entropy generated for the process is 0.09714 kJ/K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

a)

Expert Solution

Answer to Problem 215RP

The final temperature is 315.3 K.

Explanation of Solution

Write the expression for the energy balance equation for the closed system.

ΔEin−ΔEout=ΔEsystem (I)

Here, energy change in to the system is ΔEin, energy change exit from the system is ΔEout and change in energy of the system is ΔEsystem.

Write the expression for the initial pressure in the cylinder using ideal gas law.

P1=m1RT1ν1 (II)

Here, initial pressure is P1, initial mass is m1, gas constant is R, initial temperature is T1 and initial volume is ν1.

Write the expression for the mass of the air at final stage using ideal gas law.

m2=P2ν2RT2 (III)

Here, final mass is m2, final pressure is P2, final specific volume is ν2, gas constant is R, and final temperature is T2.

Write the expression for the mass entering the cylinder (mi).

mi=m2−m1 (IV)

Conclusion:

Substitute mihi for ΔEin, Wb,out for ΔEout and m2u2−m1u1 for ΔEsystem in Equation (I).

mihi−Wb,out=m2u2−m1u1micpTi−Wb,out=m2cvT2−m1cvT1 (V)

Here, work done during the process is Wb,out.mass flow is m, specific heat at constant pressure is cp , specific heat at constant volume is cv, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Substitute 1.3 kg for m1, 0.287 kJ/kg⋅K for R, 30°C for T1, and 0.40 m3 for ν1 in Equation (II).

P1=(1.3 kg)(0.287 kJ/kg⋅K)30°C0.40 m3=(1.3 kg)(0.287 kJ/kg⋅K)(30+273) K0.40 m3=282.6 kPa

When system undergoes constant pressure process, P2=P1

Substitute 282.6 kPa for P2, 1.5×0.40 m3 for ν2, 0.287 kJ/kg⋅K for R in Equation (III).

m2=(282.6 kPa)(1.5×0.40 m3)(0.287 kJ/kg⋅K)T2=590.8T2 (VI)

Substitute 590.8T2 for m2 and 1.3 kg for m1 in Equation (IV).

mi=590.8T2−1.3 kg=590.8T2−1.3 (VII)

Substitute 590.8T2−1.3 for mi, 1.005 kJ/kg⋅K for cp, 70°C for Ti, 56.52 kJ for Wb,out, 590.8T2 for m2, 0.718 kJ/kg⋅K for cv, 1.3 kg for m1 and 30°C for T1 in Equation (V).

{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70°C)−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)(30°C)}{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70+273)K−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)×(30+273)K}

(590.8T2−1.3)344.715−56.52=424.19−282.82203657.6T2−504.65=141.3742T2=203657.6646.02 =315.3 K

Thus, the final temperature is 315.3 K.

b)

To determine

The amount of mass entered the cylinder.

b)

Expert Solution

Answer to Problem 215RP

The amount of mass entered the cylinder is 0.5742 kg.

Explanation of Solution

Substitute 315.3 K for T2 in Equation (VII).

m2=590.8315.3 K=1.874 kg

Substitute 1.874 kg for m2 and 1.3 kg for m1 in Equation (V).

mi=1.874 kg−1.3 kg= 0.5742 kg

Thus, the amount of mass entered the cylinder is 0.5742 kg.

c)

To determine

The work done during the process.

c)

Expert Solution

Answer to Problem 215RP

The work done during the process is 56.52 kJ.

Explanation of Solution

Write the expression for the work done during the process (Wb,out).

Wb,out=P1(ν2−ν1) (VIII)

Here, initial volume is ν1 , final volume is ν2 and initial pressure is (P1).

Conclusion:

Substitute 282.6 kPa for P1, (150 %×0.40 m3) for ν2, and 0.40 m3 for ν1 in Equation (VIII).

Wb,out=282.6 kPa(1.5×0.40 m3−0.40 m3)=56.52 kJ

Thus, the work done during the process is 56.52 kJ.

d)

To determine

The entropy generated for the process.

d)

Expert Solution

Answer to Problem 215RP

The entropy generated for the process is 0.09714 kJ/K.

Explanation of Solution

Write the expression for the entropy generated during the process.

Sgen=m2s2−m1s1−misi=m2s2−m1s1−(m2−m1)si=m2(s2−si)−m1(s1−si)=m2(cplnT2Ti−RlnP2Pi)−m1(cplnT1Ti−RlnP1Pi) (IX)

Here, entropy generation is Sgen, initial entropy is s1,entropy entering the cylinder is si , final entropy is s2 , pressure entering the cylinder is Pi , and final pressure is P2 , temperature entering the cylinder is Ti , final temperature is T2 , gas constant is R , specific heat at constant pressure is cp, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

Conclusion:

Substitute 1.874 kg for m2, 1.005 kJ/kg⋅K for cp, 315.3 K for T2, 70 C for Ti, 0.287 kJ/kg⋅K for R, 282.6 kPa for P2, 500 kPa for Pi, 1.3 kg for m1, 30 C for T1, and 282.6 kPa for P1 in Equation (IX).

Sgen={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln30°C70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln(30+273) K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}=0.09714 kJ/K

Thus, the entropy generated for the process is 0.09714 kJ/K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

a)

Expert Solution

Answer to Problem 215RP

The final temperature is 315.3 K.

Explanation of Solution

Write the expression for the energy balance equation for the closed system.

ΔEin−ΔEout=ΔEsystem (I)

Here, energy change in to the system is ΔEin, energy change exit from the system is ΔEout and change in energy of the system is ΔEsystem.

Write the expression for the initial pressure in the cylinder using ideal gas law.

P1=m1RT1ν1 (II)

Here, initial pressure is P1, initial mass is m1, gas constant is R, initial temperature is T1 and initial volume is ν1.

Write the expression for the mass of the air at final stage using ideal gas law.

m2=P2ν2RT2 (III)

Here, final mass is m2, final pressure is P2, final specific volume is ν2, gas constant is R, and final temperature is T2.

Write the expression for the mass entering the cylinder (mi).

mi=m2−m1 (IV)

Conclusion:

Substitute mihi for ΔEin, Wb,out for ΔEout and m2u2−m1u1 for ΔEsystem in Equation (I).

mihi−Wb,out=m2u2−m1u1micpTi−Wb,out=m2cvT2−m1cvT1 (V)

Here, work done during the process is Wb,out.mass flow is m, specific heat at constant pressure is cp , specific heat at constant volume is cv, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Substitute 1.3 kg for m1, 0.287 kJ/kg⋅K for R, 30°C for T1, and 0.40 m3 for ν1 in Equation (II).

P1=(1.3 kg)(0.287 kJ/kg⋅K)30°C0.40 m3=(1.3 kg)(0.287 kJ/kg⋅K)(30+273) K0.40 m3=282.6 kPa

When system undergoes constant pressure process, P2=P1

Substitute 282.6 kPa for P2, 1.5×0.40 m3 for ν2, 0.287 kJ/kg⋅K for R in Equation (III).

m2=(282.6 kPa)(1.5×0.40 m3)(0.287 kJ/kg⋅K)T2=590.8T2 (VI)

Substitute 590.8T2 for m2 and 1.3 kg for m1 in Equation (IV).

mi=590.8T2−1.3 kg=590.8T2−1.3 (VII)

Substitute 590.8T2−1.3 for mi, 1.005 kJ/kg⋅K for cp, 70°C for Ti, 56.52 kJ for Wb,out, 590.8T2 for m2, 0.718 kJ/kg⋅K for cv, 1.3 kg for m1 and 30°C for T1 in Equation (V).

{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70°C)−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)(30°C)}{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70+273)K−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)×(30+273)K}

(590.8T2−1.3)344.715−56.52=424.19−282.82203657.6T2−504.65=141.3742T2=203657.6646.02 =315.3 K

Thus, the final temperature is 315.3 K.

b)

To determine

The amount of mass entered the cylinder.

b)

Expert Solution

Answer to Problem 215RP

The amount of mass entered the cylinder is 0.5742 kg.

Explanation of Solution

Substitute 315.3 K for T2 in Equation (VII).

m2=590.8315.3 K=1.874 kg

Substitute 1.874 kg for m2 and 1.3 kg for m1 in Equation (V).

mi=1.874 kg−1.3 kg= 0.5742 kg

Thus, the amount of mass entered the cylinder is 0.5742 kg.

c)

To determine

The work done during the process.

c)

Expert Solution

Answer to Problem 215RP

The work done during the process is 56.52 kJ.

Explanation of Solution

Write the expression for the work done during the process (Wb,out).

Wb,out=P1(ν2−ν1) (VIII)

Here, initial volume is ν1 , final volume is ν2 and initial pressure is (P1).

Conclusion:

Substitute 282.6 kPa for P1, (150 %×0.40 m3) for ν2, and 0.40 m3 for ν1 in Equation (VIII).

Wb,out=282.6 kPa(1.5×0.40 m3−0.40 m3)=56.52 kJ

Thus, the work done during the process is 56.52 kJ.

d)

To determine

The entropy generated for the process.

d)

Expert Solution

Answer to Problem 215RP

The entropy generated for the process is 0.09714 kJ/K.

Explanation of Solution

Write the expression for the entropy generated during the process.

Sgen=m2s2−m1s1−misi=m2s2−m1s1−(m2−m1)si=m2(s2−si)−m1(s1−si)=m2(cplnT2Ti−RlnP2Pi)−m1(cplnT1Ti−RlnP1Pi) (IX)

Here, entropy generation is Sgen, initial entropy is s1,entropy entering the cylinder is si , final entropy is s2 , pressure entering the cylinder is Pi , and final pressure is P2 , temperature entering the cylinder is Ti , final temperature is T2 , gas constant is R , specific heat at constant pressure is cp, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

Conclusion:

Substitute 1.874 kg for m2, 1.005 kJ/kg⋅K for cp, 315.3 K for T2, 70 C for Ti, 0.287 kJ/kg⋅K for R, 282.6 kPa for P2, 500 kPa for Pi, 1.3 kg for m1, 30 C for T1, and 282.6 kPa for P1 in Equation (IX).

Sgen={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln30°C70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln(30+273) K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}=0.09714 kJ/K

Thus, the entropy generated for the process is 0.09714 kJ/K.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

a)

Expert Solution

Answer to Problem 215RP

The final temperature is 315.3 K.

Explanation of Solution

Write the expression for the energy balance equation for the closed system.

ΔEin−ΔEout=ΔEsystem (I)

Here, energy change in to the system is ΔEin, energy change exit from the system is ΔEout and change in energy of the system is ΔEsystem.

Write the expression for the initial pressure in the cylinder using ideal gas law.

P1=m1RT1ν1 (II)

Here, initial pressure is P1, initial mass is m1, gas constant is R, initial temperature is T1 and initial volume is ν1.

Write the expression for the mass of the air at final stage using ideal gas law.

m2=P2ν2RT2 (III)

Here, final mass is m2, final pressure is P2, final specific volume is ν2, gas constant is R, and final temperature is T2.

Write the expression for the mass entering the cylinder (mi).

mi=m2−m1 (IV)

Conclusion:

Substitute mihi for ΔEin, Wb,out for ΔEout and m2u2−m1u1 for ΔEsystem in Equation (I).

mihi−Wb,out=m2u2−m1u1micpTi−Wb,out=m2cvT2−m1cvT1 (V)

Here, work done during the process is Wb,out.mass flow is m, specific heat at constant pressure is cp , specific heat at constant volume is cv, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Substitute 1.3 kg for m1, 0.287 kJ/kg⋅K for R, 30°C for T1, and 0.40 m3 for ν1 in Equation (II).

P1=(1.3 kg)(0.287 kJ/kg⋅K)30°C0.40 m3=(1.3 kg)(0.287 kJ/kg⋅K)(30+273) K0.40 m3=282.6 kPa

When system undergoes constant pressure process, P2=P1

Substitute 282.6 kPa for P2, 1.5×0.40 m3 for ν2, 0.287 kJ/kg⋅K for R in Equation (III).

m2=(282.6 kPa)(1.5×0.40 m3)(0.287 kJ/kg⋅K)T2=590.8T2 (VI)

Substitute 590.8T2 for m2 and 1.3 kg for m1 in Equation (IV).

mi=590.8T2−1.3 kg=590.8T2−1.3 (VII)

Substitute 590.8T2−1.3 for mi, 1.005 kJ/kg⋅K for cp, 70°C for Ti, 56.52 kJ for Wb,out, 590.8T2 for m2, 0.718 kJ/kg⋅K for cv, 1.3 kg for m1 and 30°C for T1 in Equation (V).

{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70°C)−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)(30°C)}{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70+273)K−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)×(30+273)K}

(590.8T2−1.3)344.715−56.52=424.19−282.82203657.6T2−504.65=141.3742T2=203657.6646.02 =315.3 K

Thus, the final temperature is 315.3 K.

b)

To determine

The amount of mass entered the cylinder.

b)

Expert Solution

Answer to Problem 215RP

The amount of mass entered the cylinder is 0.5742 kg.

Explanation of Solution

Substitute 315.3 K for T2 in Equation (VII).

m2=590.8315.3 K=1.874 kg

Substitute 1.874 kg for m2 and 1.3 kg for m1 in Equation (V).

mi=1.874 kg−1.3 kg= 0.5742 kg

Thus, the amount of mass entered the cylinder is 0.5742 kg.

c)

To determine

The work done during the process.

c)

Expert Solution

Answer to Problem 215RP

The work done during the process is 56.52 kJ.

Explanation of Solution

Write the expression for the work done during the process (Wb,out).

Wb,out=P1(ν2−ν1) (VIII)

Here, initial volume is ν1 , final volume is ν2 and initial pressure is (P1).

Conclusion:

Substitute 282.6 kPa for P1, (150 %×0.40 m3) for ν2, and 0.40 m3 for ν1 in Equation (VIII).

Wb,out=282.6 kPa(1.5×0.40 m3−0.40 m3)=56.52 kJ

Thus, the work done during the process is 56.52 kJ.

d)

To determine

The entropy generated for the process.

d)

Expert Solution

Answer to Problem 215RP

The entropy generated for the process is 0.09714 kJ/K.

Explanation of Solution

Write the expression for the entropy generated during the process.

Sgen=m2s2−m1s1−misi=m2s2−m1s1−(m2−m1)si=m2(s2−si)−m1(s1−si)=m2(cplnT2Ti−RlnP2Pi)−m1(cplnT1Ti−RlnP1Pi) (IX)

Here, entropy generation is Sgen, initial entropy is s1,entropy entering the cylinder is si , final entropy is s2 , pressure entering the cylinder is Pi , and final pressure is P2 , temperature entering the cylinder is Ti , final temperature is T2 , gas constant is R , specific heat at constant pressure is cp, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

Conclusion:

Substitute 1.874 kg for m2, 1.005 kJ/kg⋅K for cp, 315.3 K for T2, 70 C for Ti, 0.287 kJ/kg⋅K for R, 282.6 kPa for P2, 500 kPa for Pi, 1.3 kg for m1, 30 C for T1, and 282.6 kPa for P1 in Equation (IX).

Sgen={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln30°C70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln(30+273) K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}=0.09714 kJ/K

Thus, the entropy generated for the process is 0.09714 kJ/K.

Answer 6

Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

a)

Expert Solution

Answer to Problem 215RP

The final temperature is 315.3 K.

Explanation of Solution

Write the expression for the energy balance equation for the closed system.

ΔEin−ΔEout=ΔEsystem (I)

Here, energy change in to the system is ΔEin, energy change exit from the system is ΔEout and change in energy of the system is ΔEsystem.

Write the expression for the initial pressure in the cylinder using ideal gas law.

P1=m1RT1ν1 (II)

Here, initial pressure is P1, initial mass is m1, gas constant is R, initial temperature is T1 and initial volume is ν1.

Write the expression for the mass of the air at final stage using ideal gas law.

m2=P2ν2RT2 (III)

Here, final mass is m2, final pressure is P2, final specific volume is ν2, gas constant is R, and final temperature is T2.

Write the expression for the mass entering the cylinder (mi).

mi=m2−m1 (IV)

Conclusion:

Substitute mihi for ΔEin, Wb,out for ΔEout and m2u2−m1u1 for ΔEsystem in Equation (I).

mihi−Wb,out=m2u2−m1u1micpTi−Wb,out=m2cvT2−m1cvT1 (V)

Here, work done during the process is Wb,out.mass flow is m, specific heat at constant pressure is cp , specific heat at constant volume is cv, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Substitute 1.3 kg for m1, 0.287 kJ/kg⋅K for R, 30°C for T1, and 0.40 m3 for ν1 in Equation (II).

P1=(1.3 kg)(0.287 kJ/kg⋅K)30°C0.40 m3=(1.3 kg)(0.287 kJ/kg⋅K)(30+273) K0.40 m3=282.6 kPa

When system undergoes constant pressure process, P2=P1

Substitute 282.6 kPa for P2, 1.5×0.40 m3 for ν2, 0.287 kJ/kg⋅K for R in Equation (III).

m2=(282.6 kPa)(1.5×0.40 m3)(0.287 kJ/kg⋅K)T2=590.8T2 (VI)

Substitute 590.8T2 for m2 and 1.3 kg for m1 in Equation (IV).

mi=590.8T2−1.3 kg=590.8T2−1.3 (VII)

Substitute 590.8T2−1.3 for mi, 1.005 kJ/kg⋅K for cp, 70°C for Ti, 56.52 kJ for Wb,out, 590.8T2 for m2, 0.718 kJ/kg⋅K for cv, 1.3 kg for m1 and 30°C for T1 in Equation (V).

{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70°C)−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)(30°C)}{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70+273)K−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)×(30+273)K}

(590.8T2−1.3)344.715−56.52=424.19−282.82203657.6T2−504.65=141.3742T2=203657.6646.02 =315.3 K

Thus, the final temperature is 315.3 K.

Answer 7

Chapter 7.13, Problem 215RP

a)

To determine

The final temperature.

Answer 8

a)

Expert Solution

Answer to Problem 215RP

The final temperature is 315.3 K.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the energy balance equation for the closed system.

ΔEin−ΔEout=ΔEsystem (I)

Here, energy change in to the system is ΔEin, energy change exit from the system is ΔEout and change in energy of the system is ΔEsystem.

Write the expression for the initial pressure in the cylinder using ideal gas law.

P1=m1RT1ν1 (II)

Here, initial pressure is P1, initial mass is m1, gas constant is R, initial temperature is T1 and initial volume is ν1.

Write the expression for the mass of the air at final stage using ideal gas law.

m2=P2ν2RT2 (III)

Here, final mass is m2, final pressure is P2, final specific volume is ν2, gas constant is R, and final temperature is T2.

Write the expression for the mass entering the cylinder (mi).

mi=m2−m1 (IV)

Conclusion:

Substitute mihi for ΔEin, Wb,out for ΔEout and m2u2−m1u1 for ΔEsystem in Equation (I).

mihi−Wb,out=m2u2−m1u1micpTi−Wb,out=m2cvT2−m1cvT1 (V)

Here, work done during the process is Wb,out.mass flow is m, specific heat at constant pressure is cp , specific heat at constant volume is cv, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

From the Table A-2” Ideal-gas specific heats of various common gases” obtain the following properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Substitute 1.3 kg for m1, 0.287 kJ/kg⋅K for R, 30°C for T1, and 0.40 m3 for ν1 in Equation (II).

P1=(1.3 kg)(0.287 kJ/kg⋅K)30°C0.40 m3=(1.3 kg)(0.287 kJ/kg⋅K)(30+273) K0.40 m3=282.6 kPa

When system undergoes constant pressure process, P2=P1

Substitute 282.6 kPa for P2, 1.5×0.40 m3 for ν2, 0.287 kJ/kg⋅K for R in Equation (III).

m2=(282.6 kPa)(1.5×0.40 m3)(0.287 kJ/kg⋅K)T2=590.8T2 (VI)

Substitute 590.8T2 for m2 and 1.3 kg for m1 in Equation (IV).

mi=590.8T2−1.3 kg=590.8T2−1.3 (VII)

Substitute 590.8T2−1.3 for mi, 1.005 kJ/kg⋅K for cp, 70°C for Ti, 56.52 kJ for Wb,out, 590.8T2 for m2, 0.718 kJ/kg⋅K for cv, 1.3 kg for m1 and 30°C for T1 in Equation (V).

{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70°C)−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)(30°C)}{(590.8T2−1.3)(1.005 kJ/kg⋅K)×(70+273)K−56.52 kJ}={(590.8T2)(0.718 kJ/kg⋅K)T2−(1.3 kg)×(0.718 kJ/kg⋅K)×(30+273)K}

(590.8T2−1.3)344.715−56.52=424.19−282.82203657.6T2−504.65=141.3742T2=203657.6646.02 =315.3 K

Thus, the final temperature is 315.3 K.

Answer 13

b)

To determine

The amount of mass entered the cylinder.

b)

Expert Solution

Answer to Problem 215RP

The amount of mass entered the cylinder is 0.5742 kg.

Explanation of Solution

Substitute 315.3 K for T2 in Equation (VII).

m2=590.8315.3 K=1.874 kg

Substitute 1.874 kg for m2 and 1.3 kg for m1 in Equation (V).

mi=1.874 kg−1.3 kg= 0.5742 kg

Thus, the amount of mass entered the cylinder is 0.5742 kg.

Answer 14

b)

To determine

The amount of mass entered the cylinder.

Answer 15

b)

Expert Solution

Answer to Problem 215RP

The amount of mass entered the cylinder is 0.5742 kg.

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Substitute 315.3 K for T2 in Equation (VII).

m2=590.8315.3 K=1.874 kg

Substitute 1.874 kg for m2 and 1.3 kg for m1 in Equation (V).

mi=1.874 kg−1.3 kg= 0.5742 kg

Thus, the amount of mass entered the cylinder is 0.5742 kg.

Answer 20

c)

To determine

The work done during the process.

c)

Expert Solution

Answer to Problem 215RP

The work done during the process is 56.52 kJ.

Explanation of Solution

Write the expression for the work done during the process (Wb,out).

Wb,out=P1(ν2−ν1) (VIII)

Here, initial volume is ν1 , final volume is ν2 and initial pressure is (P1).

Conclusion:

Substitute 282.6 kPa for P1, (150 %×0.40 m3) for ν2, and 0.40 m3 for ν1 in Equation (VIII).

Wb,out=282.6 kPa(1.5×0.40 m3−0.40 m3)=56.52 kJ

Thus, the work done during the process is 56.52 kJ.

Answer 21

c)

To determine

The work done during the process.

Answer 22

c)

Expert Solution

Answer to Problem 215RP

The work done during the process is 56.52 kJ.

Answer 23

c)

Expert Solution

Answer 24

c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for the work done during the process (Wb,out).

Wb,out=P1(ν2−ν1) (VIII)

Here, initial volume is ν1 , final volume is ν2 and initial pressure is (P1).

Conclusion:

Substitute 282.6 kPa for P1, (150 %×0.40 m3) for ν2, and 0.40 m3 for ν1 in Equation (VIII).

Wb,out=282.6 kPa(1.5×0.40 m3−0.40 m3)=56.52 kJ

Thus, the work done during the process is 56.52 kJ.

Answer 27

d)

To determine

The entropy generated for the process.

d)

Expert Solution

Answer to Problem 215RP

The entropy generated for the process is 0.09714 kJ/K.

Explanation of Solution

Write the expression for the entropy generated during the process.

Sgen=m2s2−m1s1−misi=m2s2−m1s1−(m2−m1)si=m2(s2−si)−m1(s1−si)=m2(cplnT2Ti−RlnP2Pi)−m1(cplnT1Ti−RlnP1Pi) (IX)

Here, entropy generation is Sgen, initial entropy is s1,entropy entering the cylinder is si , final entropy is s2 , pressure entering the cylinder is Pi , and final pressure is P2 , temperature entering the cylinder is Ti , final temperature is T2 , gas constant is R , specific heat at constant pressure is cp, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

Conclusion:

Substitute 1.874 kg for m2, 1.005 kJ/kg⋅K for cp, 315.3 K for T2, 70 C for Ti, 0.287 kJ/kg⋅K for R, 282.6 kPa for P2, 500 kPa for Pi, 1.3 kg for m1, 30 C for T1, and 282.6 kPa for P1 in Equation (IX).

Sgen={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln30°C70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln(30+273) K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}=0.09714 kJ/K

Thus, the entropy generated for the process is 0.09714 kJ/K.

Answer 28

d)

To determine

The entropy generated for the process.

Answer 29

d)

Expert Solution

Answer to Problem 215RP

The entropy generated for the process is 0.09714 kJ/K.

Answer 30

d)

Expert Solution

Answer 31

d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the expression for the entropy generated during the process.

Sgen=m2s2−m1s1−misi=m2s2−m1s1−(m2−m1)si=m2(s2−si)−m1(s1−si)=m2(cplnT2Ti−RlnP2Pi)−m1(cplnT1Ti−RlnP1Pi) (IX)

Here, entropy generation is Sgen, initial entropy is s1,entropy entering the cylinder is si , final entropy is s2 , pressure entering the cylinder is Pi , and final pressure is P2 , temperature entering the cylinder is Ti , final temperature is T2 , gas constant is R , specific heat at constant pressure is cp, mass entering the cylinder is mi, final mass is m2 and initial mass is m2.

Conclusion:

Substitute 1.874 kg for m2, 1.005 kJ/kg⋅K for cp, 315.3 K for T2, 70 C for Ti, 0.287 kJ/kg⋅K for R, 282.6 kPa for P2, 500 kPa for Pi, 1.3 kg for m1, 30 C for T1, and 282.6 kPa for P1 in Equation (IX).

Sgen={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln30°C70°C−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}={(1.874 kg)((1.005 kJ/kg⋅K)ln315.3 K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)−(1.3 kg)((1.005 kJ/kg⋅K)ln(30+273) K(70+273) K−0.287 kJ/kg⋅Kln282.6 kPa500 kPa)}=0.09714 kJ/K